I think you might be mistaken. If $R^2=r^2$, I assume you have a bivariate model: one DV, one IV. I don't think $R^2$ will change if you swap these, nor if you replace the IV with the predictions of the DV that are based on the IV. Here's code for a demonstration in R:
x=rnorm(1000); y=rnorm(1000) #store random data
summary(lm(y~x)) #fit a linear regression model (a)
summary(lm(x~y)) #swap variables and fit the opposite model (b)
z=lm(y~x)$fitted.values; summary(lm(y~z)) #substitute predictions for the IV in model a
x=rnorm(1000); y=rnorm(1000) # store random data
summary(lm(y~x)) # fit a linear regression model (a)
summary(lm(x~y)) # swap variables and fit the opposite model (b)
z=lm(y~x)$fitted.values; summary(lm(y~z)) # substitute predictions for IV in model (a)
If you aren't working with a bivariate model, your choice of DV will affect $R^2$...unless your variables are all identically correlated, I suppose, but this isn't much of an exception. If all the variables have identical strengths of correlation and also share the same portions of the DV's variance (e.g. [or maybe "i.e."], if some of the variables are completely identical), you could just reduce this to a bivariate model without losing any information. Whether you do or don't, $R^2$ still wouldn't change.
In all other cases I can think of with more than two variables, $R^2\ne r^2$ where $R^2$ is the coefficient of determination and $r$ is a bivariate correlation coefficient of any kind (not necessarily Pearson's; e.g., possibly also a Spearman's $\rho$).
